If the digits of a 3 digit number abc is reversed, then the difference of original and reversed number is not divisible by 9.

True

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False

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Solution

The correct option is B
False

Let the number be abc.
abc = 100a + 10b + c
After reversing the digits: cba = 100c + 10b + a

abc - cba = (100a + 10b + c) - (100c + 10b + a)
= 100a + 10b + c - 100c - 10b - a
= 100a + c - 100c - a
= 100a - a - 100c + c
= 99a - 99c
= 99(a - c)

∴ $99 (a - c) = 9 \times 11 (a - c)$

$D i v i d e n d = d i v i s o r \times q u o t i e n t + r e m a i n d e r$

$99 (a - c) = 9 \times [11 \times (a - c)] + r e m a i n d e r$

Remainder = 0; so the number is divisible by 9. Hence the statement is incorrect.

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